Expand Using the Binomial Theorem (2k^2+1)^2

${(2 k^{2} + 1)}^{2}$

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(2 k^{2})}^{2 - k} \cdot {(1)}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} \cdot {(2 k^{2})}^{2 - 0} \cdot {(1)}^{0} + \frac{2!}{(2 - 1)! 1!} \cdot {(2 k^{2})}^{2 - 1} \cdot {(1)}^{1} + \frac{2!}{(2 - 2)! 2!} \cdot {(2 k^{2})}^{2 - 2} \cdot {(1)}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 k^{2})}^{2} \cdot {(1)}^{0} + 2 \cdot {(2 k^{2})}^{1} \cdot {(1)}^{1} + 1 \cdot {(2 k^{2})}^{0} \cdot {(1)}^{2}$

Step 4

Simplify each term.

Multiply $1$ by ${(1)}^{0}$ by adding the exponents.

Move ${(1)}^{0}$ .

${(1)}^{0} \cdot 1 \cdot {(2 k^{2})}^{2} + 2 \cdot {(2 k^{2})}^{1} \cdot {(1)}^{1} + 1 \cdot {(2 k^{2})}^{0} \cdot {(1)}^{2}$

Multiply ${(1)}^{0}$ by $1$ .

Raise $1$ to the power of $1$ .

${(1)}^{0} \cdot 1^{1} \cdot {(2 k^{2})}^{2} + 2 \cdot {(2 k^{2})}^{1} \cdot {(1)}^{1} + 1 \cdot {(2 k^{2})}^{0} \cdot {(1)}^{2}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$1^{0 + 1} \cdot {(2 k^{2})}^{2} + 2 \cdot {(2 k^{2})}^{1} \cdot {(1)}^{1} + 1 \cdot {(2 k^{2})}^{0} \cdot {(1)}^{2}$

Add $0$ and $1$ .

$1^{1} \cdot {(2 k^{2})}^{2} + 2 \cdot {(2 k^{2})}^{1} \cdot {(1)}^{1} + 1 \cdot {(2 k^{2})}^{0} \cdot {(1)}^{2}$

Simplify $1^{1} \cdot {(2 k^{2})}^{2}$ .

${(2 k^{2})}^{2} + 2 \cdot {(2 k^{2})}^{1} \cdot {(1)}^{1} + 1 \cdot {(2 k^{2})}^{0} \cdot {(1)}^{2}$

Apply the product rule to $2 k^{2}$ .

$2^{2} {(k^{2})}^{2} + 2 \cdot {(2 k^{2})}^{1} \cdot {(1)}^{1} + 1 \cdot {(2 k^{2})}^{0} \cdot {(1)}^{2}$

Raise $2$ to the power of $2$ .

$4 {(k^{2})}^{2} + 2 \cdot {(2 k^{2})}^{1} \cdot {(1)}^{1} + 1 \cdot {(2 k^{2})}^{0} \cdot {(1)}^{2}$

Multiply the exponents in ${(k^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4 k^{2 \cdot 2} + 2 \cdot {(2 k^{2})}^{1} \cdot {(1)}^{1} + 1 \cdot {(2 k^{2})}^{0} \cdot {(1)}^{2}$

Multiply $2$ by $2$ .

$4 k^{4} + 2 \cdot {(2 k^{2})}^{1} \cdot {(1)}^{1} + 1 \cdot {(2 k^{2})}^{0} \cdot {(1)}^{2}$

Simplify.

$4 k^{4} + 2 \cdot (2 k^{2}) \cdot {(1)}^{1} + 1 \cdot {(2 k^{2})}^{0} \cdot {(1)}^{2}$

Multiply $2$ by $2$ .

$4 k^{4} + 4 k^{2} \cdot {(1)}^{1} + 1 \cdot {(2 k^{2})}^{0} \cdot {(1)}^{2}$

Evaluate the exponent.

$4 k^{4} + 4 k^{2} \cdot 1 + 1 \cdot {(2 k^{2})}^{0} \cdot {(1)}^{2}$

Multiply $4$ by $1$ .

$4 k^{4} + 4 k^{2} + 1 \cdot {(2 k^{2})}^{0} \cdot {(1)}^{2}$

Multiply $1$ by ${(1)}^{2}$ by adding the exponents.

Move ${(1)}^{2}$ .

$4 k^{4} + 4 k^{2} + {(1)}^{2} \cdot 1 \cdot {(2 k^{2})}^{0}$

Multiply ${(1)}^{2}$ by $1$ .

Raise $1$ to the power of $1$ .

$4 k^{4} + 4 k^{2} + {(1)}^{2} \cdot 1^{1} \cdot {(2 k^{2})}^{0}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$4 k^{4} + 4 k^{2} + 1^{2 + 1} \cdot {(2 k^{2})}^{0}$

Add $2$ and $1$ .

$4 k^{4} + 4 k^{2} + 1^{3} \cdot {(2 k^{2})}^{0}$

Simplify $1^{3} \cdot {(2 k^{2})}^{0}$ .

$4 k^{4} + 4 k^{2} + 1^{3}$

One to any power is one.

$4 k^{4} + 4 k^{2} + 1$

Expand Using the Binomial Theorem (2k^2+1)^2

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